The segre cubic primal $X\subset P^4$ is the GIT quotient of 6 points on $P^1$. Let $M_{0,6}$ the DM compactification of the moduli of 6-pointed rational curves. The Segre primal $X$ is a cubic 3-fold with ten double points and there exists a natural map $M_{0,6}\to X$ that contracts 10 boundary divisors (each iso to $P^1 \times P^1$) to the singular points. Now if I blow up the singular points of $X$, since they are ordinary double points, I get a $P^1 \times P^1$ exceptional divisor over each of them. Call $\tilde{X}$ the blown up variety. Maybe it is a silly question, but it is clear that there exists an iso $\tilde{X}\cong M_{0,6}$? Why?

2 Answers
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From what you've written you'll get a map $M_{0,6} \to \tilde{X}$ via the universal property of blowups (see Proposition II.7.14 of Hartshorne's Algebraic Geometry). In fact this map is an isomorphism.

As Steven said, there is a morphism $\overline M_{0,6} \to \tilde{X}$ because the inverse image sheaf of the ideal of double points generates a Cartier divisor.
Now $\tilde{X}$ is nonsingular, so to check the fact $\overline{M}_{0,6} \to \tilde{X}$ is an isomorphism it suffices to show injectivity of the map, or show the equality of Picard numbers.

In general, the birational morphism $\overline M_{0,n} \to (\mathbb{P}^1)^n//SL_2$ is a composition of smooth blow-ups and Kirwan's desingularization. Consult http://arxiv.org/abs/1002.2461.

yes I see. is the fact the inverse image sheaf of the ideal of the blow up is locally free equivalent to the schematic inverse image being smooth and locally complete intersection?
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IMeasyJan 28 '13 at 8:49

no, probably it is wrong: think of the ideal sheaf of the origin in $\mathbb{A}^2$. that's a counterexample.
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IMeasyJan 28 '13 at 8:50